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4 years ago

8

A circular hoop sits in a stream of water, oriented perpendicular to the current. If the area of the hoop is doubled, the flux (

volume of water per unit time) through it:___________

1 answer:

natka813 [3]4 years ago

3 0

Answer:

The flux (volume of water per unit time) through the hoop will also double.

Explanation:

The flux = volume of water per unit time = flow rate of water through the hoop.

The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.

This means that

Flow rate = AV

where A is the area of the hoop

V is the velocity of the water through the hoop

This flow rate = volume of water per unit time = Δv/Δt =Q

From all the above statements, we can say

Q = AV

From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2

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Select the correct answer. Which existing technology did engineers use to enhance the speed of propeller-driven airplanes

Musya8 [376]

metallurgy:

Explanation:

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2 years ago

At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =

Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

$F=ma$ for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that $W = mg$

For simplicity we work with $g =9.807$ $\frac{m}{s^{2}}$ despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula $g = a-bz$ the "normal" or "standard" weight of an object is given by $W = mg = ma$ when $z = 0$ , so we need to find the value of $z$ that makes $W = m(a-bz) = 0.997ma$ meaning that the original weight decrease by a 0.30%, so now we operate...

$m(a-bz) = 0.997ma$ now we group like terms on the same sides $ma(1-0.997) = mbz$ we cancel equal tems on both sides and obtain that $z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m$

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3 years ago

5 cause of a electrical problem

cupoosta [38]

Answer:

1. Poor circuit protection

2.Grounding issue

3. lighting problem

4. Electrical shocks

5. High electricity bills

Explanation:

8 0

3 years ago

The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc

a_sh-v [17]

Answer:

(a)

<em>d</em>Q = m<em>d</em>q

<em>d</em>q = $C_p$ <em>d</em>T

$q = \int\limits^{T_2}_{T_1} {C_p} \, dT$ = $C_p$ (T₂ - T₁)

From the above equations, the underlying assumption is that $C_p$ remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ , Q₂ = 6.14 KJ

ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter

Let $C_{cal}$ be heat constant of calorimeter

Q₂ = $C_{cal}$ ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m $C_p$ ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³ at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m $C_p$ ' ΔT

$C_p$ = (16.73 - 6.14) / (15.085 x 3.10)

$C_p$ = 0.22646 KJ mol⁻¹ k⁻¹

6 0

3 years ago

What’s the number of gold atoms in a nanogram? a picogram?

zvonat [6]

Answer :

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

The number of gold atoms in picogram is, $3.057\times 10^{9}$

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains $6.022\times 10^{23}$ number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-9}$ nanograms of gold contains $\frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12}$ number of gold atoms

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-12}$ picograms of gold contains $\frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9}$ number of gold atoms

The number of gold atoms in picogram is, $3.057\times 10^{9}$

8 0

3 years ago